A liquid of density p is coming out of a hose pipe of radius a with horizontal speed `upsilon` and hits a mesh . 50 % of the liquid passes through the mesh unaffected . 25 % comes back with the same speed .The resultant pressure on the mesh will be:

Options

A. `1//4 upsilon^2`
B. `3//4 upsilon^2`
C. `1//2 upsilon^2`
D. `upsilon^2`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Density of the liquid = \( \rho \)
- Radius of the hose pipe = \( a \)
- Horizontal speed of the liquid = \( v \)

2. **Calculate the Volume Flow Rate**:
The volume flow rate \( Q \) of the liquid coming out of the hose pipe can be calculated using the formula:
\[ Q = A \cdot v \]
where \( A \) is the cross-sectional area of the hose pipe, given by:
\[ A = \pi a^2 \]
Thus, the volume flow rate becomes:
\[ Q = \pi a^2 v \]

3. **Determine the Mass Flow Rate**:
The mass flow rate \( \dot{m} \) can be calculated as:
\[ \dot{m} = \rho Q = \rho (\pi a^2 v) \]

4. **Calculate Forces Acting on the Mesh**:
- **Force due to liquid passing through the mesh**: 50% of the liquid passes through unaffected. Therefore, the mass flow rate through the mesh is:
\[ \dot{m}_{\text{through}} = 0.5 \dot{m} = 0.5 \rho (\pi a^2 v) \]
The momentum change (force) due to this portion is:
\[ F_1 = \dot{m}_{\text{through}} \cdot v = 0.5 \rho (\pi a^2 v) \cdot v = 0.5 \rho \pi a^2 v^2 \]

- **Force due to liquid coming back**: 25% of the liquid comes back with the same speed. The mass flow rate for this portion is:
\[ \dot{m}_{\text{back}} = 0.25 \dot{m} = 0.25 \rho (\pi a^2 v) \]
The momentum change (force) due to this portion is:
\[ F_2 = \dot{m}_{\text{back}} \cdot v = 0.25 \rho (\pi a^2 v) \cdot v = 0.25 \rho \pi a^2 v^2 \]

5. **Calculate the Total Force on the Mesh**:
The total force \( F \) on the mesh is the sum of the forces from both portions:
\[ F = F_1 + F_2 = 0.5 \rho \pi a^2 v^2 + 0.25 \rho \pi a^2 v^2 = 0.75 \rho \pi a^2 v^2 \]

6. **Calculate the Pressure on the Mesh**:
The pressure \( P \) on the mesh can be calculated by dividing the total force by the area of the mesh \( A \):
\[ P = \frac{F}{A} \]
Assuming the area of the mesh is \( A = \pi a^2 \), we have:
\[ P = \frac{0.75 \rho \pi a^2 v^2}{\pi a^2} = 0.75 \rho v^2 \]

7. **Final Result**:
The resultant pressure on the mesh is:
\[ P = \frac{3}{4} \rho v^2 \]